Question: Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{q^2 - 16}{q - 4}$
Solution: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = q$ $ b = \sqrt{16} = -4$ So we can rewrite the expression as: $a = \dfrac{({q} {-4})({q} + {4})} {q - 4} $ We can divide the numerator and denominator by $(q - 4)$ on condition that $q \neq 4$ Therefore $a = q + 4; q \neq 4$